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Homework Help: Proportion Problems

  1. Nov 11, 2004 #1
    This has been racking my brain for awhile, and I know I have it right, but something is missing.

    1. A person earns 850 euros in one week. How much will she earn in 3.5 weeks?

    This seemed simple enough, so I just multiplied 850 by 3.5 and got $2,975. I know there's one more step, and I can't find it.

    2. A picture is 73/4 inches wide and 81/2 inches high. If you wish to enlarge the width to 10 inches, how high will the picture be? Express in fractions instead of decimals.

    I converted the fractions into decimals so I can get the fraction answer. I came up with 7.75/8.5 = x/10 = I then got 77.5 over 8.5 and came up with 911/100. Obviously this is wrong, but I can't figure out the step I missed here.

    3. There's 5/4 of a triangle and 2/x of another one, I'm supposed to find the missing piece.I come up with 8/5. I'm stuck on where to go from here.

    Any help would be appericated.
     
  2. jcsd
  3. Nov 11, 2004 #2

    Doc Al

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    Staff: Mentor

    This one's just fine---no missing steps.

    You messed up the statement of ratios. It looks like you tried to use:
    [tex]\frac{W_1}{H_1} = \frac{W_2}{H_2}[/tex]
    But you put [itex]H_2 = 10[/itex] instead of [itex]W_2 = 10[/itex].
    I don't understand this problem.
     
  4. Nov 11, 2004 #3

    NateTG

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    Right. You forgot to account for taxes ;). Just kidding, that looks correct to me.

    I'd suggest staying with fractions if you're comforatable:

    So you've got an inital aspect ratio:
    [tex]\frac{width}{height}=\frac{\frac{73}{4}}{\frac{81}{2}}=\frac{146}{324}=\frac{73}{162}[/tex]
    and then you want so solve for the same aspect ratio with a different width:
    [tex]\frac{10}{h}=\frac{73}{162}[/tex]
    [tex]10\times162=73 h[/tex]
    [tex]\frac{1620}{73}=h[/tex]

    Of course, since [tex]\frac{73}{4} > 10[/tex] and the problem mentions increase to 10 inches, I'm guessing you copied it incorrectly.

    I can't quite follow what you're doing, but I'd guess that you'should be getting [tex]\frac{8}{5x}[/tex] instead of [tex]\frac{8}{5}[/tex].
     
  5. Nov 11, 2004 #4
    The exact problem is

    I have two triangles, one is 5 inches on the left and 4 inches at the bottom. Another one is 2 inches on the left side and x on the bottom. The idea is to find the missing piece of that one, that's what I was trying to say.
     
  6. Nov 11, 2004 #5

    NateTG

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    Can you fill in the spaces below?
    5 is to 4 as _ is to _
     
  7. Nov 11, 2004 #6
    5 is to 4 as 2 is to x.

    One triange is 5 inches on the left, and 4 at the bottom. Another is 2 inches to the left. The idea is to find the x at the bottom of the other triange.
     
  8. Nov 11, 2004 #7
    Let me redo this one:

    A picture is 7 3/4 inches wide and is 8 1/2 inches high. If you enlarge the picture by 10 inches, how high will the picture be expressed as a fraction over a decimal?
     
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